We consider the system AotoYamada_05__013.

  Alphabet:

    append : [c * c] --> c 
    cons : [b * c] --> c 
    flatwith : [a -> b * b] --> c 
    flatwithsub : [a -> b * c] --> c 
    leaf : [a] --> b 
    nil : [] --> c 
    node : [c] --> b 

  Rules:

    append(nil, x) => x 
    append(cons(x, y), z) => cons(x, append(y, z)) 
    flatwith(f, leaf(x)) => cons(f x, nil) 
    flatwith(f, node(x)) => flatwithsub(f, x) 
    flatwithsub(f, nil) => nil 
    flatwithsub(f, cons(x, y)) => append(flatwith(f, x), flatwithsub(f, y)) 

This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).

We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]).

We thus obtain the following dependency pair problem (P_0, R_0, static, formative):

  Dependency Pairs P_0:

    0] append#(cons(X, Y), Z) =#> append#(Y, Z)   
    1] flatwith#(F, node(X)) =#> flatwithsub#(F, X)   
    2] flatwithsub#(F, cons(X, Y)) =#> append#(flatwith(F, X), flatwithsub(F, Y))   
    3] flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
    4] flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, Y)   

  Rules R_0:

    append(nil, X) => X 
    append(cons(X, Y), Z) => cons(X, append(Y, Z)) 
    flatwith(F, leaf(X)) => cons(F X, nil) 
    flatwith(F, node(X)) => flatwithsub(F, X) 
    flatwithsub(F, nil) => nil 
    flatwithsub(F, cons(X, Y)) => append(flatwith(F, X), flatwithsub(F, Y)) 

Thus, the original system is terminating if (P_0, R_0, static, formative) is finite.

We consider the dependency pair problem (P_0, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  append#(cons(X, Y), Z) >? append#(Y, Z) 
  flatwith#(F, node(X)) >? flatwithsub#(F, X) 
  flatwithsub#(F, cons(X, Y)) >? append#(flatwith(F, X), flatwithsub(F, Y)) 
  flatwithsub#(F, cons(X, Y)) >? flatwith#(F, X) 
  flatwithsub#(F, cons(X, Y)) >? flatwithsub#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  flatwith(F, leaf(X)) >= cons(F X, nil) 
  flatwith(F, node(X)) >= flatwithsub(F, X) 
  flatwithsub(F, nil) >= nil 
  flatwithsub(F, cons(X, Y)) >= append(flatwith(F, X), flatwithsub(F, Y)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  append = \y0y1.y1 
  append# = \y0y1.0 
  cons = \y0y1.0 
  flatwith = \G0y1.0 
  flatwith# = \G0y1.3 
  flatwithsub = \G0y1.0 
  flatwithsub# = \G0y1.3 
  leaf = \y0.3 + 2y0 
  nil = 0 
  node = \y0.3 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[append#(_x1, _x2)]] 
  [[flatwith#(_F0, node(_x1))]] = 3 >= 3 = [[flatwithsub#(_F0, _x1)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 3 > 0 = [[append#(flatwith(_F0, _x1), flatwithsub(_F0, _x2))]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 3 >= 3 = [[flatwith#(_F0, _x1)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 3 >= 3 = [[flatwithsub#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = x2 >= 0 = [[cons(_x0, append(_x1, _x2))]] 
  [[flatwith(_F0, leaf(_x1))]] = 0 >= 0 = [[cons(_F0 _x1, nil)]] 
  [[flatwith(_F0, node(_x1))]] = 0 >= 0 = [[flatwithsub(_F0, _x1)]] 
  [[flatwithsub(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[flatwithsub(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[append(flatwith(_F0, _x1), flatwithsub(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_0, R_0, static, formative) by (P_1, R_0, static, formative), where P_1 consists of:

  append#(cons(X, Y), Z) =#> append#(Y, Z)   
  flatwith#(F, node(X)) =#> flatwithsub#(F, X)   
  flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
  flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, Y)   

Thus, the original system is terminating if (P_1, R_0, static, formative) is finite.

We consider the dependency pair problem (P_1, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  append#(cons(X, Y), Z) >? append#(Y, Z) 
  flatwith#(F, node(X)) >? flatwithsub#(F, X) 
  flatwithsub#(F, cons(X, Y)) >? flatwith#(F, X) 
  flatwithsub#(F, cons(X, Y)) >? flatwithsub#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  flatwith(F, leaf(X)) >= cons(F X, nil) 
  flatwith(F, node(X)) >= flatwithsub(F, X) 
  flatwithsub(F, nil) >= nil 
  flatwithsub(F, cons(X, Y)) >= append(flatwith(F, X), flatwithsub(F, Y)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  append = \y0y1.y0 + y1 
  append# = \y0y1.0 
  cons = \y0y1.y0 + y1 
  flatwith = \G0y1.y1G0(y1) 
  flatwith# = \G0y1.2y1 
  flatwithsub = \G0y1.2y1G0(y1) 
  flatwithsub# = \G0y1.2y1 
  leaf = \y0.3 + y0 
  nil = 0 
  node = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[append#(_x1, _x2)]] 
  [[flatwith#(_F0, node(_x1))]] = 6 + 4x1 > 2x1 = [[flatwithsub#(_F0, _x1)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 2x1 + 2x2 >= 2x1 = [[flatwith#(_F0, _x1)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 2x1 + 2x2 >= 2x2 = [[flatwithsub#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 
  [[flatwith(_F0, leaf(_x1))]] = 3F0(3 + x1) + x1F0(3 + x1) >= F0(x1) = [[cons(_F0 _x1, nil)]] 
  [[flatwith(_F0, node(_x1))]] = 2x1F0(3 + 2x1) + 3F0(3 + 2x1) >= 2x1F0(x1) = [[flatwithsub(_F0, _x1)]] 
  [[flatwithsub(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[flatwithsub(_F0, cons(_x1, _x2))]] = 2x1F0(x1 + x2) + 2x2F0(x1 + x2) >= 2x2F0(x2) + x1F0(x1) = [[append(flatwith(_F0, _x1), flatwithsub(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_1, R_0, static, formative) by (P_2, R_0, static, formative), where P_2 consists of:

  append#(cons(X, Y), Z) =#> append#(Y, Z)   
  flatwithsub#(F, cons(X, Y)) =#> flatwith#(F, X)   
  flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, Y)   

Thus, the original system is terminating if (P_2, R_0, static, formative) is finite.

We consider the dependency pair problem (P_2, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  append#(cons(X, Y), Z) >? append#(Y, Z) 
  flatwithsub#(F, cons(X, Y)) >? flatwith#(F, X) 
  flatwithsub#(F, cons(X, Y)) >? flatwithsub#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  flatwith(F, leaf(X)) >= cons(F X, nil) 
  flatwith(F, node(X)) >= flatwithsub(F, X) 
  flatwithsub(F, nil) >= nil 
  flatwithsub(F, cons(X, Y)) >= append(flatwith(F, X), flatwithsub(F, Y)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  append = \y0y1.y1 
  append# = \y0y1.0 
  cons = \y0y1.0 
  flatwith = \G0y1.1 + y1 + 2y1G0(y1) 
  flatwith# = \G0y1.0 
  flatwithsub = \G0y1.0 
  flatwithsub# = \G0y1.3 
  leaf = \y0.3 
  nil = 0 
  node = \y0.3 + y0 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 0 >= 0 = [[append#(_x1, _x2)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 3 > 0 = [[flatwith#(_F0, _x1)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 3 >= 3 = [[flatwithsub#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = x2 >= 0 = [[cons(_x0, append(_x1, _x2))]] 
  [[flatwith(_F0, leaf(_x1))]] = 4 + 6F0(3) >= 0 = [[cons(_F0 _x1, nil)]] 
  [[flatwith(_F0, node(_x1))]] = 4 + x1 + 2x1F0(3 + x1) + 6F0(3 + x1) >= 0 = [[flatwithsub(_F0, _x1)]] 
  [[flatwithsub(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[flatwithsub(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[append(flatwith(_F0, _x1), flatwithsub(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_2, R_0, static, formative) by (P_3, R_0, static, formative), where P_3 consists of:

  append#(cons(X, Y), Z) =#> append#(Y, Z)   
  flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, Y)   

Thus, the original system is terminating if (P_3, R_0, static, formative) is finite.

We consider the dependency pair problem (P_3, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  append#(cons(X, Y), Z) >? append#(Y, Z) 
  flatwithsub#(F, cons(X, Y)) >? flatwithsub#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  flatwith(F, leaf(X)) >= cons(F X, nil) 
  flatwith(F, node(X)) >= flatwithsub(F, X) 
  flatwithsub(F, nil) >= nil 
  flatwithsub(F, cons(X, Y)) >= append(flatwith(F, X), flatwithsub(F, Y)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  append = \y0y1.y0 + y1 
  append# = \y0y1.y0 
  cons = \y0y1.2 + y1 + 3y0 
  flatwith = \G0y1.3y1 + 2y1G0(y1) 
  flatwithsub = \G0y1.y1 + y1G0(y1) 
  flatwithsub# = \G0y1.0 
  leaf = \y0.3 + y0 
  nil = 0 
  node = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[append#(cons(_x0, _x1), _x2)]] = 2 + x1 + 3x0 > x1 = [[append#(_x1, _x2)]] 
  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 0 >= 0 = [[flatwithsub#(_F0, _x2)]] 
  [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 2 + x1 + x2 + 3x0 >= 2 + x1 + x2 + 3x0 = [[cons(_x0, append(_x1, _x2))]] 
  [[flatwith(_F0, leaf(_x1))]] = 9 + 3x1 + 2x1F0(3 + x1) + 6F0(3 + x1) >= 2 + 3F0(x1) = [[cons(_F0 _x1, nil)]] 
  [[flatwith(_F0, node(_x1))]] = 9 + 6x1 + 4x1F0(3 + 2x1) + 6F0(3 + 2x1) >= x1 + x1F0(x1) = [[flatwithsub(_F0, _x1)]] 
  [[flatwithsub(_F0, nil)]] = 0 >= 0 = [[nil]] 
  [[flatwithsub(_F0, cons(_x1, _x2))]] = 2 + x2 + 3x1 + 2F0(2 + x2 + 3x1) + 3x1F0(2 + x2 + 3x1) + x2F0(2 + x2 + 3x1) >= x2 + 3x1 + 2x1F0(x1) + x2F0(x2) = [[append(flatwith(_F0, _x1), flatwithsub(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace the dependency pair problem (P_3, R_0, static, formative) by (P_4, R_0, static, formative), where P_4 consists of:

  flatwithsub#(F, cons(X, Y)) =#> flatwithsub#(F, Y)   

Thus, the original system is terminating if (P_4, R_0, static, formative) is finite.

We consider the dependency pair problem (P_4, R_0, static, formative).

We will use the reduction pair processor [Kop12, Thm. 7.16].  It suffices to find a standard reduction pair [Kop12, Def. 6.69].  Thus, we must orient:

  flatwithsub#(F, cons(X, Y)) >? flatwithsub#(F, Y) 
  append(nil, X) >= X 
  append(cons(X, Y), Z) >= cons(X, append(Y, Z)) 
  flatwith(F, leaf(X)) >= cons(F X, nil) 
  flatwith(F, node(X)) >= flatwithsub(F, X) 
  flatwithsub(F, nil) >= nil 
  flatwithsub(F, cons(X, Y)) >= append(flatwith(F, X), flatwithsub(F, Y)) 

We orient these requirements with a polynomial interpretation in the natural numbers.

The following interpretation satisfies the requirements:

  append = \y0y1.y0 + y1 
  cons = \y0y1.1 + y1 + 2y0 
  flatwith = \G0y1.3y1 + y1G0(y1) 
  flatwithsub = \G0y1.2y1 + 2y1G0(y1) 
  flatwithsub# = \G0y1.y1 
  leaf = \y0.3 + y0 
  nil = 1 
  node = \y0.3 + 2y0 

Using this interpretation, the requirements translate to:

  [[flatwithsub#(_F0, cons(_x1, _x2))]] = 1 + x2 + 2x1 > x2 = [[flatwithsub#(_F0, _x2)]] 
  [[append(nil, _x0)]] = 1 + x0 >= x0 = [[_x0]] 
  [[append(cons(_x0, _x1), _x2)]] = 1 + x1 + x2 + 2x0 >= 1 + x1 + x2 + 2x0 = [[cons(_x0, append(_x1, _x2))]] 
  [[flatwith(_F0, leaf(_x1))]] = 9 + 3x1 + 3F0(3 + x1) + x1F0(3 + x1) >= 2 + 2F0(x1) = [[cons(_F0 _x1, nil)]] 
  [[flatwith(_F0, node(_x1))]] = 9 + 6x1 + 2x1F0(3 + 2x1) + 3F0(3 + 2x1) >= 2x1 + 2x1F0(x1) = [[flatwithsub(_F0, _x1)]] 
  [[flatwithsub(_F0, nil)]] = 2 + 2F0(1) >= 1 = [[nil]] 
  [[flatwithsub(_F0, cons(_x1, _x2))]] = 2 + 2x2 + 4x1 + 2x2F0(1 + x2 + 2x1) + 2F0(1 + x2 + 2x1) + 4x1F0(1 + x2 + 2x1) >= 2x2 + 3x1 + 2x2F0(x2) + x1F0(x1) = [[append(flatwith(_F0, _x1), flatwithsub(_F0, _x2))]] 

By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_4, R_0) by ({}, R_0).  By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed.

As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.


+++ Citations +++

[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
[Kop13]  C. Kop.  Static Dependency Pairs with Accessibility.  Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013.
[KusIsoSakBla09]  K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui.  Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems.  In volume 92(10) of IEICE Transactions on Information and Systems.  2007--2015, 2009.